How many grams of PbCl2 can be formed from creating a 100mL (0.5M) solution of Pb(NO3)2 and a 100mL (0.5M) solution of CaCl2?

Question

I have this problem for my Chemistry 1111 lecture that I'm stuck on and I was wondering if someone could help me out, any sort of help would be appreciated, thanks.

Answer 1

Jay, calculate the moles of Pb(2+): that's molarity of the Pb(2+) solution x moles (Pb)/mole Pb-salt [this is 1/1] x volume of that solution. Then calculate the moles of Cl(sub2): that's molarity of the CaCl2 solution x moles Cl(sub2)/mole Ca-salt [that is 1/1] x volume of that solution.

In this case, you have an exact stoichiometric match of Pb to Cl(sub2), so theoretically you could recover that number of moles of PbCl2. [Note that if you didn't have an exact match, one ingredient would be in excess, and the other would be limiting. In that case, you would use the limiting one for your calculations -- you can't make any more of it (unless you are a star, perhaps in nova stage) -- and the excess of the other material is left over.] Multiply by the molar mass of PbCl2, and you have the result.

In practical terms, if you were trying to *recover* the PbCl2 as a solid precipitate (your problem didn't say this, I'm just giving you some additional advice for the future time when you might have to consider this), you would then do a calculation of the solubility of PbCl2, using a value called the *solubility product* of PbCl2. The amount of soluble PbCl2 you calculate you would then subtract from the total PbCl2 present, to find the amount of PbCl2 expected to precipitate as a solid.

 

Just commented back and the browser wiped it out ... oh well.
OK.
For Pb(2+): 0.5 mol/L x 0.1 L x 1/1 = 0.05 mol
For Cl2: 0.5 mol/L x 0.1 L x 1/1 = 0.05 mol of Cl2
So together they make 0.05 mol PbCl2.
0.05 mol PbCl2 x (207.2 g/mol Pb + 2 x 35.453 g/mol Cl) = 0.05 mol x 278.106 g/mol = 13.9053 g
Now, if your instructor wants you to consider all the numbers in the problem as exact, you could round to 13.91 g because the lead MW has 4 significant digits and the Cl MW has 5 significant digits.
Myself, I would tend to regard these as only as significant as they're given, and round to 14 g (strictly speaking, the "0.5" only has 1 significant figure, but significant digits are only a way of approximating the uncertainties of precision of numbers....).
now some practical thoughts: If you measured your liquids out with a graduated cylinder, the "100 mL" might be to +/- 0.5 mL; and who knows about the 0.5 M solution! (But if you measured with a Class A TD (to deliver) volumetric pipet, you should be able to do better than this -- maybe +/- 0.1 mL.)